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I bought a pack of tealights each powered by a single CR2032 (flickering LED; builtin timer turns it on every evening for 5 h). My wife loves them, but after just a fortnight each tealight's CR2032 has dropped to 2.2 V (under load), and the LED has become invisibly dim. Discharge curves suggest that the mean load is 2.5 mA (which a CR2032 can maintain for 70 h, yielding 170 mAh).

Because I'd rather not replace several CR2032's per week, I'd like to replace the CR2032 with a pair of alkaline AA's.

AA's have roughly 10x the mAh, but in published discharge curves for AA's, the lowest testing current I found was forty times higher than my load.

Logging data for over a month would take unusual patience. Has anyone done so?

To be precise: at a 2.5 mA load, how many hours will it take for an alkaline AA (of whatever brand name) to discharge down to 1.1 V? (So a pair, down to 2.2 V, to emulate the CR2032.)

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  • \$\begingroup\$ An Alkaline AA starts at 1.5 volts or so, so you'll need two in series to replace the CR2032s. \$\endgroup\$ – Peter Bennett May 2 '16 at 19:26
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Typical Alkaline AA is 2000-2300mAh. Self discharge is around 2-3% per year or 60mAh.

I've tested 100 mA discharge rate to 1.1V, yielding around 1800mAh. For so much lower than this, I would expect you get very close to nominal capacity. This would be around 800 hours.

This is assuming your mean load doesn't include high spikes. Alkaline internal resistance is around 0.15 ohms. Lithium coins cells are around 0.01-0.04 ohms. So their losses for short spikes will be a little larger than the coin cell (and overall run time decreased slightly). If the current draw is fairly constant, then this isn't a factor.

Ideal system would be using Eneloop NiMh cells for this. They are generally around or below 1.2V under "normal" use with higher current. With this light of a current, they will hover around 1.3V for most of the discharge and get around 1900mAh before dropping through 1.1V. Self discharge over the month would drop this to an effective 1800mAh or so. This should give you around 30 days of use and a recharge to get back going. Best option for both the pocket book and environment.

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If the lowest testing current you found was forty times higher than your load, take the time from the curve and multiply it by forty, you'll have the time you can stand with your load. Actually, it will certainly be significantly more than that, because, at lower current, the internal resistance of the battery has less effect. So you'll gain more mAh.

Here is what I found: some curve shown me that a AA battery can stand 7h at 250mA (to 1.1V). Other curves I have found shown that it can stand 18 to 20h at 100mA (depending on the model). That is consistent: 7h*250mA/100mA = 17.5h. Now, at 2.5 mA, it means you can stand at least 18h*100mA/2.5mA = 720h. And actually, it will be more than that because of the resistance thing. Maybe 800h. Something in that range, depending on the brand/model of battery.

Curve sources:

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  • \$\begingroup\$ So just extrapolate from the 250mA and 100mA curves to estimate a 2.5mA curve. I like it. \$\endgroup\$ – Camille Goudeseune May 2 '16 at 19:49
  • \$\begingroup\$ Exactly. At high current, the curve may show different results (less capacity) because of the internal resistance, as I said. But for low currents, the effects are less dramatic, so you just extrapolate. \$\endgroup\$ – dim May 2 '16 at 19:53
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    \$\begingroup\$ If the time spans are very long, one also need to consider self discharge though. I don't think it will matter in this application, but it is good to know in my opinion. \$\endgroup\$ – caconyrn May 2 '16 at 20:00

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